For C-C bonds, the order of length (shortest to longest) is

Answer: C≡C, C=C, C-C Higher bond order → more attraction → shorter bond. Triple < double < single.

For a given pair of atoms, bond strength generally

Answer: Increases as bond order increases (single < double < triple) Higher bond order = more shared electron density = stronger and shorter bond.

What does this trend show about bond order?

Answer: As bond order increases, bond length decreases Higher bond order pulls atoms closer; triple bonds are shorter than double, which are shorter than single bonds.

AP Chemistry · Topic 2.2

Intramolecular Force and Potential Energy Practice

Q: For a given pair of atoms, bond strength generally

Answer: Increases as bond order increases (single < double < triple) Higher bond order = more shared electron density = stronger and shorter bond.

Q: What does this trend show about bond order?

Answer: As bond order increases, bond length decreases Higher bond order pulls atoms closer; triple bonds are shorter than double, which are shorter than single bonds.

Part of Compound Structure and Properties.(SAP-3.B)

Practice questions

Want a predicted score for the whole AP CHEM exam? Take the 20-question diagnostic and Lumi will plan the rest.

Take diagnosticarrow_forward Practice all 2.2

Sample questions

4 of 4 — sign in to practice the rest with adaptive difficulty and mastery tracking.

Sample 1difficulty 2/5
$For C-C bonds, the order of length (shortest to longest) is$
- A
  $C\equivC, C=C, C-C$
  check_circle
- B
  $C=C, C\equivC, C-C$
- C
  $C-C, C=C, C\equivC$
- D
  $All same$
Why
$Higher bond order \to more attraction \to shorter bond. Triple < double < single.$
View full solution & statsarrow_forward
Sample 2difficulty 2/5
$For a given pair of atoms, bond strength generally$
- A
  $Is unrelated to bond order$
- B
  $Is the same regardless$
- C
  $Increases as bond order increases (single < double < triple)$
  check_circle
- D
  $Decreases as bond order increases$
Why
$Higher bond order = more shared electron density = stronger and shorter bond.$
View full solution & statsarrow_forward
Sample 3difficulty 3/5
$What does this trend show about bond order?$
- A
  $As bond order increases, bond length decreases$
  check_circle
- B
  $As bond order increases, bond length increases$
- C
  $Bonds of order 1.5 are longest$
- D
  $Bond order is unrelated to length$
Why
$Higher bond order pulls atoms closer; triple bonds are shorter than double, which are shorter than single bonds.$
View full solution & statsarrow_forward
Sample 4difficulty 3/5
$The trend C-C < C=C < C\equivC in bond energy implies:$
- A
  $Higher bond order means longer, weaker bonds$
- B
  $Higher bond order means shorter, stronger bonds$
  check_circle
- C
  $Bond order does not affect bond strength$
- D
  $Triple bonds have only sigma overlap$
Why
$Multiple bonds bring atoms closer and concentrate more electron density between them, raising bond energy.$
View full solution & statsarrow_forward

Predict my AP scorearrow_forward